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TL431: How to calculate power of series resistor

Denish Limbachiya
Prodigy 30 points
Part Number: TL431


Tool/software:

Hello Team

in our one of the design we utilize TL431BQDBZRQ1 shunt regulator 

We want to create 10V from Vin range ( 13V to 35V)

please find an attached sch for the review.

Query 1: Can you check and confirm power calculation of R6 series resistor is correct as per mansion in note? And selected R6 series resistor 's power (1W) is enough for desire operation?

  • 0
    Prodigy 30 points

    Is that any simulation tool or Evalution file is available for the TL431BQDBZRQ1 shunt regulator?

  • 0 in reply to Denish Limbachiya
    TI__Expert 4710 points

    Hi Denish,

    Thanks for your question!

    The power dissipation by the TL431 would be:

    Pz_max = Vz * Iz = 10V * 0.075A = 0.75W

    This is the case with no load where all current goes through the cathode of the TL431.

    Is the load current expected to be 8mA?

    If so, the power dissipation over the resistor R6 would be as follows:

    P_R = I_R * V_R = 75.7mA * (35V-10V) = 1.9W

    Additionally, please find the output capacitor stability chart below and adjust C6 so that it is in the stable region for Vka = 10V as in your application:

    I hope this helps!

    Best Regards,

    Andrew Li

  • 0 in reply to Andrew Li
    Prodigy 30 points

    Hello Andrew,

    Yes, our Iout = 8mA

    So according to that 

    Series Resistor
    Power P(R(Series) = (Vcc-Vz) * Iz
    = (35-10) * 0.008
    = 0.2W = 200mW

    is it ok?

    We have replaced C5 and C6 1Uf cap with 2.2uF cap for the safer margin and have removed C4

    Is it ok for our design?

    Please confirm above two things

    Thanks,

    Denish Limbachiya

  • 0 in reply to Denish Limbachiya
    TI__Expert 4710 points

    Hi Denish,

    Thanks for the update.

    1.

    For the power of the series resistor (R6) the maximum current situation should be considered. When VDC_IN = 35V, the current through the resistor would be I_R = (35V - 10V)/330Ohm = 75.7mA.

    As a result, P_R = I_R * V_R = 75.7mA * (35V-10V) = 1.9W

    Although the load may only draw 8mA, the rest of the current will be shunted through the TL431. The resistor R6 will have all of this current pass through it.

    2. 

    The capacitor values look fine to me.

    Best Regards,

    Andrew Li

  • 0 in reply to Andrew Li
    Prodigy 30 points

    Hello Andrew

    Thanks for your valuable inputs.

  • +1 in reply to Denish Limbachiya
    TI__Expert 4710 points

    Hi Denish,

    I'm glad to hear this helped!

    Since there are no further questions, I will close this thread. To reopen this thread, simply reply back to this message.

    Best Regards,

    Andrew Li